We\'re going to titrate formic acid with the strong There is initially 100. mL o
ID: 972350 • Letter: W
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We're going to titrate formic acid with the strong There is initially 100. mL of 0.50 M formic acid and the concentration of NaOH is 10 M. All work must be shown to receive credit. The of We're going to formic acid with the strong basic, NaOH. There is initially 100 mL of 0.50 M formic acid and the concentration of NaOH is 10 M. All work must be shown to receive credit. The quiz will be due at the beginning of class on 4/20 16. What is the initial pH of the formic acid solution? What is the percent ionization under initial conditions? After the addition of 10 ml. of NaOH. what is the pH? After the addition of 25 ml. of NaOH. what is the pH? Think about where in the titration this brings you.Explanation / Answer
HCOOH + NaOH HCOO- + Na+ + H2O
From the equation we can see that the ratio between HCOOH and NaOH is 1:1
Amt of HCOOH we started of with = 0.1*0.5= 0.05 mol
1. Before any NaOH is added (0.00mL) there is only HCOOH present.
pH = -log[HCOONa]
pH = -log(0.5)
pH = 0.301
2) HCOOH = H+ + HCOO-
Ka of HCOOH = 1.8 x 10e-4 = [H+][HCOO-]/ [HCOOH] = c * (a^2), where a= ionisation const
a= 0.06
So percentage of ionisation = 6 %.
3. When 10mL (0.01L) of 1M NaOH is added we are creating a mixture of acid and base
The pH can be determined by the use of the Henderson-Hasselbalch equation:
pH = -log(Ka) + log{[HA]/[A-]}
where:
[HA] = the conc. of the base
[A-] = the conc. of the acid
n(NaOH) = CV
n(NaOH) = 1 x 0.01 L
n(NaOH) = 0.01 moles
Out of the 0.05 mol of HCOOH that we started with … 0.01 moles is reacts with the NaOH .Therefore:
0.05 moles of HCOOH – 0.01moles
= 0.04moles of HCOOH left!!
Now we determine the concentration of the solution, remember the volume has INCREASED due to the fact that we have just added 10mL of NaOH to the 50mL of HCOOH, therefore the new volume is 60ml (0.06L),
C(HCOOH) = n / V
C(HCOOH) = 0.04 / 0.11
C(HCOOH) = 0.363M
Therefore we can now plug the information into the equation:
pH = -log(Ka(HCOOH)) + log{[HA]/[A-]}
pH = -log(1.8 x 10e-4) + log (0.01/0.363)
pH = 2.19
4) 25mL of 1 M NaOH
n(NaOH) = CV
n(NaOH) = 1 x 0.025 L
n(NaOH) = 0.025 moles
0.05 moles of HCOOH – 0.025moles
= 0.025moles of HCOOH left!!
the volume has INCREASED due to the fact that we have just added an extra 15mL of NaOH on top of the 10mL so now we have added 25mL of NaOH to the 50mL of HCOOH, therefore the new volume is 75ml (0.075L),
C(HCOOH) = n / V
C(HCOOH) = 0.025 / 0.075
C(HCOOH) = 0.333
Again, we can now plug the information into the equation:
pH = -log(Ka) + log{[HA]/[A-]}
pH = -log(1.8 x 10e-4) + log (0.025/0.333)
pH = 2.62
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